# Homework Help: Limits on an Integral of a semi-circle

1. May 28, 2012

### ZedCar

1. The problem statement, all variables and given/known data

A question asks to calculate the integral over the region R given by:

x^2 + y^2 <= 4
0 <= y <= 2

Which would be the upper half of a circle of radius 2 centred on the origin.

The integral is done in the book I have and the limits of x are given as -2 to 2, which I can understand.

Though the limits for y are given as: 0 to (4 - x^2)^0.5

I can see that they have obtained this limit from rearranging the first part of the region R.

BUT, why is the limit for y not 0 to 2. Or alternatively, if what they have done is correct, why is it not equally valid to state the limits for x are: 0 to (4 - y^2)^0.5

2. Relevant equations

3. The attempt at a solution

2. May 28, 2012

### algebrat

When integrating over a region, we need to hit each point in the region exactly once, and no other points.

So your recipe is, for each x betwen -2 and 2, (hold x fixed and) integrate over y=0 to 2. But then you would hit all points in a rectangle, not the semicircle.

You'll hit each point in the semicircle once if, for each x between -2 and 2, you go from y= 0 to the upper circle.

If this seemed to make sense, here's some exercises to check that you understand.

Try switching order of integration, that is, for each y in between (which values?), let x runf from where to where, thus hitting each point in the region.

Another exercise. In the original limits, hold x fixed and let y vary, why do we take x goes from -2 to 2. Why not a smaller or larger region, what exactly would be wrong with that, how geometrically or by what type of value would it change the answer. (Hint: integration works if we hit each point in region exactly once, and no others.)

3. May 28, 2012

### tiny-tim

Hi ZedCar!
It is, if the y limits are -2 to 2.

You can have vertical slices of thickness dx and height √(4 - x2)

or horizontal slices of thickness dy and width √(4 - y2).